1. Find $f(x)$

Given $f'(x)=f(x), \quad f(0)=1.$

The differential equation $f'(x)=f(x)$ has solution $f(x)=Ce^x.$ Using $f(0)=1$: $1=Ce^0=C \implies C=1.$ Hence, $f(x)=e^x.$

2. Find $g(x)$

We are given $f(x)+g(x)=x^2.$ So, $g(x)=x^2-f(x)=x^2-e^x.$

3. Form the product $f(x)g(x)$

$f(x)g(x)=e^x(x^2-e^x)=x^2e^x-e^{2x}.$

Therefore, $\int_0^1 f(x)g(x)\,dx=\int_0^1 x^2e^x\,dx-\int_0^1 e^{2x}\,dx.$

4. Evaluate $\int_0^1 x^2e^x\,dx$

Use the standard result (or integration by parts twice): $\int x^2e^x\,dx=e^x(x^2-2x+2)+C.$

So, $\int_0^1 x^2e^x\,dx=\left[e^x(x^2-2x+2)\right]_0^1.$

At $x=1$: $e^1(1-2+2)=e.$ At $x=0$: $e^0(0-0+2)=2.$ Thus, $\int_0^1 x^2e^x\,dx=e-2.$

5. Evaluate $\int_0^1 e^{2x}\,dx$

$\int_0^1 e^{2x}\,dx=\left[\frac{e^{2x}}{2}\right]_0^1=\frac{e^2-1}{2}.$

6. Compute the required integral

$\int_0^1 f(x)g(x)\,dx=(e-2)-\frac{e^2-1}{2}.$

Simplify: $=e-2-\frac{e^2}{2}+\frac{1}{2}$ $=e-\frac{e^2}{2}-\frac{3}{2}.$

7. Match with the options

This is $e-\frac{e^2}{2}-\frac{3}{2},$ which corresponds to Option D.